Can u pls help me with this question

Put brackets in the calculation below to make it correct.<br><br> 15 - 4 x 2 + 1 = 3

bagirrra123 [75]

If you put brackets around (2+1), your method of working is:
1) 15-4*(2+1)=3
2) 15-4*3=3
3) 15-12=3
You don't need any more brackets, as the BIDMAS (brackets, Indices, division, multiplication, addition, subtraction) rule does the rest of the job for you.
The answer is therefore: 15-4*(2+1)=3

3 0

3 years ago

line passes through the point (0, –1) and has a positive slope. Which of these points could that line pass through? Check all th

sammy [17]

Answer:

Points passing though are (12, 3),(–2, –5) and (1, 15)

Step-by-step explanation:

For a line passing through (x₁,y₁) and (x₂,y₂) slope is given by $\frac{y_2-y_1}{x_2-x_1}$ .

Here we have one point (0, –1), we need to check slope of all the points given.

(12, 3)

$m=\frac{3-(-1))}{12-0}=\frac{1}{4}>0$

(–2, –5)

$m=\frac{-5-(-1))}{-2-0}=\frac{-4}{-2}=2>0$

(–3, 1)

$m=\frac{1-(-1))}{-3-0}=\frac{2}{-3}=-\frac{2}{3}$

(1, 15)

$m=\frac{15-(-1))}{1-0}=\frac{16}{1}=16>0$

(5, –2)

$m=\frac{-2-(-1))}{5-0}=\frac{-1}{5}=-\frac{1}{5}$

Points with positive slope are (12, 3),(–2, –5) and (1, 15)

8 0

4 years ago

Read 2 more answers

I need help please, I am so confused.

vazorg [7]

Answer:

$\begin{tabular}{| c | c | c |}\cline{1-3} Equation & x-intercepts & x-coordinate of vertex\\\cline{1-3} $y=x(x-2)$ & $x=0, x=2$ & $x=1$\\\cline{1-3} $y=(x-4)(x+5)$ & $x=-5, x=4$ & $x=-0.5$\\\cline{1-3} $y=-5x(x-3)$ & $x=0, x=3$ & $x=1.5$\\\cline{1-3} \end{tabular}$

Step-by-step explanation:

x-intercepts are when the curve intercepts the x-axis, so when y =0.

Therefore, to find the x-intercepts, substitute y = 0 and solve for x.

The vertex is the turning point: the minimum point of a parabola that opens upward, and the maximum point of the parabola that opens downward. As a parabola is symmetrical, the x-coordinate of the vertex is the midpoint of the x-intercepts.

Equation: $y=x(x-2)$

$\implies x(x-2)=0$

$\implies x=0$

$\implies (x-2)=0 \implies x=2$

Therefore, the x-intercepts are x = 0 and x = 2

The midpoint of the x-intercepts is x = 1, so the x-coordinate of the vertex is x = 1

Equation: $y=(x-4)(x+5)$

$\implies (x-4)(x+5)=0$

$\implies (x-4)=0 \implies x=4$

$\implies (x+5)=0 \implies x=-5$

Therefore, the x-intercepts are x = -5 and x = 4

The midpoint of the x-intercepts is x = -0.5, so the x-coordinate of the vertex is x = -0.5

Equation: $y=-5x(3-x)$

$\implies -5x(3-x)=0$

$\implies -5x=0 \implies x=0$

$\implies (3-x)=0 \implies x=3$

Therefore, the x-intercepts are x = 0 and x = 3

The midpoint of the x-intercepts is x = 1.5, so the x-coordinate of the vertex is x = 1.5

$\begin{tabular}{| c | c | c |}\cline{1-3} Equation & x-intercepts & x-coordinate of vertex\\\cline{1-3} $y=x(x-2)$ & $x=0, x=2$ & $x=1$\\\cline{1-3} $y=(x-4)(x+5)$ & $x=-5, x=4$ & $x=-0.5$\\\cline{1-3} $y=-5x(x-3)$ & $x=0, x=3$ & $x=1.5$\\\cline{1-3} \end{tabular}$

4 0

2 years ago

Could anyone find this?

LiRa [457]

A and B are correct for this one!!